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What is sign of y if ball is thrown up?

  1. Sep 29, 2015 #1
    This is a general question that i am confused about. If I throw a ball up, then is the distance that it reaches at the very top the y or y-initial? Also, is this y or y-initial positive or negative?
     
  2. jcsd
  3. Sep 29, 2015 #2

    berkeman

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    Staff: Mentor

    You can define your coordinates however is the most convenient for solving any particular problem. The simplest 2-d coordinate system for projectile motion problems would be to put the origin on the ground, with the x-axis horizontal and the y-axis vertical. That means y=0 is at ground level, so if you are 6 feet tall and throw a ball up, then y(t=0)=6', and Vy(0)= whatever velocity you release the ball at time t=0. The height at the top of the ball's travel will depend on Vy(0) at the release point y(0)=6' plus however long it takes for the ball to stop moving up. Use an equation like this...

    y(t) = y(0) + Vy(0)*t + 1/2*ay*t^2

    Where a = the acceleration due to gravity, which in this case is -9.8m/s^2.

    Does that help?
     
  4. Sep 29, 2015 #3
    Yes, thank you
     
  5. Sep 29, 2015 #4
    Also, the time that the ball takes to come down is twice the time the ball takes to reach its maximum height, right?
     
  6. Sep 29, 2015 #5

    berkeman

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    Not if you release it from your 6' height. Right? :smile:
     
  7. Sep 29, 2015 #6

    berkeman

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    Staff: Mentor

    Well, I mean the two times from release to apex, and from apex to hitting the ground. Those are not equal.
     
  8. Sep 29, 2015 #7
    Yea, in understand it now thank you.
     
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